Simplify the following expression: $y = \dfrac{4x^2- 1x- 18}{4x - 9}$
First use factoring by grouping to factor the expression in the numerator. This expression is in the form ${A}x^2 + {B}x + {C}$ First, find two values, $a$ and $b$ , so: $ \begin{eqnarray} {ab} &=& {A}{C} \\ {a} + {b} &=& {B} \end{eqnarray} $ In this case: $ \begin{eqnarray} {ab} &=& {(4)}{(-18)} &=& -72 \\ {a} + {b} &=& &=& {-1} \end{eqnarray} $ In order to find ${a}$ and ${b}$ , list out the factors of $-72$ and add them together. Remember, since $-72$ is negative, one of the factors must be negative. The factors that add up to ${-1}$ will be your ${a}$ and ${b}$ When ${a}$ is ${-9}$ and ${b}$ is ${8}$ $ \begin{eqnarray} {ab} &=& ({-9})({8}) &=& -72 \\ {a} + {b} &=& {-9} + {8} &=& -1 \end{eqnarray} $ Next, rewrite the expression as $({A}x^2 + {a}x) + ({b}x + {C})$ $ ({4}x^2 {-9}x) + ({8}x {-18}) $ Factor out the common factors: $ x(4x - 9) + 2(4x - 9)$ Now factor out $(4x - 9)$ $ (4x - 9)(x + 2)$ The original expression can therefore be written: $ \dfrac{(4x - 9)(x + 2)}{4x - 9}$ We are dividing by $4x - 9$ , so $4x - 9 \neq 0$ Therefore, $x \neq \frac{9}{4}$ This leaves us with $x + 2; x \neq \frac{9}{4}$.